10.1 In this example, both $540 and $80 are statistics.
10.2 In this example 68% is a parameter and 73% is a statistic.
10.3 In this example 2.5003cm is a parameter and 2.5009cm is a statistic.
10.5 It would be unwise to sell only 12 policies because in the unlikely event that one of them required a payout, the insurance company would not have nearly enough money to pay. Selling thousands of policies is good business because with so many policies, it is more likely that the proportion of claims will be near the expected number and there are enough other homeowners who have paid in that there should be enough money to cover the claims made.
10.7
a) The standard deviation of Juan's mean result is 10/1.73 = 5.77.
b) Juan must repeat the experiment 4 times to reduce the standard deviation to 5. The average of a number of measurements is more likely to be close to the true value than a single measurement.
10.9
a) If X is the score of a random 12th-grader, then P(X > 300) =.5 since 300 is the mean of the distribution. P(X > 335)=P(Z > 1)= .1587.
b) The mean score of 4 12th-graders selected at random has a Normal distribution with mean 300 and standard deviation 35/2. The probability that the mean score of 4 12th-graders is higher than 300 is .5. The probability that the mean score of 4 12th-graders is higher than 335 is .0228.
10.10 The approximate probability that the average loss is greater than $275 is found by computing the z-score: (275-250)/(1000/100) = 2.5. Looking this up in the Table A gives a left tail probability of .9938, so the right tail (or the probability of the average being greater than 275) is 1-.9938 = .0062.
10.17 In this example, the sample mean 65 is a statistic. The population mean 64 is a parameter.
10.18 The law of large numbers implies that over many years, Joe will probably have won back about 60 cents for each dollar he spent.
10.20
a) The distribution from samples of size 100 is much closer to the Normal distribution. The central limit theorem implies that the sample average gets closer to a Normal distribution as the sample size increases.
b) The range of sample average values for sample size 25 appears to be (20,64). The range of sample average values for sample size 100 appears to be (28, 50). The standard deviation of the sample average decreases as sample size increases.
c) The mean income of the population appears to be about $38,000.
10.21 With a sample size of 500, we can use the normal approximation for the distribution of x to find the probability that the average fee paid by the sample households exceeds $29. z= (29-28)/(10/22.36) = 2.236. Using Table A, the probability that the average fee paid by the sample households exceeds $29 is 0.0125.
10.24
a) (140-125)/10 = 1.5, so the probability that Shelia is diagnosed as having gestational diabetes is 0.0668.
b) (140-125)/(10/2) = 3.0, so the probability of being diagnosed with gestational diabetes is .0013.
10.27
a) The approximate distribution of x is Normal with a mean of 2.2 and a standard deviation of 1.4/7.2 = 0.19.
b) (2-2.2)/.19 = -1.05, so the probability that x<2 is approximately .1469.
c) 100 accidents in a year would correspond to an average of 100/52= 1.92 accidents per week. (1.92-2.2)/.19 =-1.47, so the probability that there are fewer than 100 accidents in a year is approximately .0708.
10.28
a) The distribution cannot be Normal since the possible values are discrete and the Normal distribution is continuous.
b) The approximate distribution of x is Normal with a mean of 1.5 and standard deviation .028.
c) If 700 cars carry 1075 people, that means there is an average of 1.54 people per car. The z-score for 1.54 is (1.54-1.5)/.028=1.43, so the probability of the average being greater than 1.54 (and therefore the total being greater than 1075) is .0764.
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